hdu 5459

Problem Description

I've sent Fang Fang around 201314 text messages in almost 5 years. Why can't she make sense of what I mean?
``But Jesus is here!" the priest intoned. ``Show me your messages."
Fine, the first message is s1=‘‘c" and the second one is s2=‘‘ff".
The i-th message is si=si−2+si−1 afterwards. Let me give you some examples.
s3=‘‘cff", s4=‘‘ffcff" and s5=‘‘cffffcff".

``I found the i-th message's utterly charming," Jesus said.
``Look at the fifth message". s5=‘‘cffffcff" and two ‘‘cff" appear in it.
The distance between the first ‘‘cff" and the second one we said, is 5.
``You are right, my friend," Jesus said. ``Love is patient, love is kind.
It does not envy, it does not boast, it is not proud. It does not dishonor others, it is not self-seeking, it is not easily angered, it keeps no record of wrongs.
Love does not delight in evil but rejoices with the truth.
It always protects, always trusts, always hopes, always perseveres."

Listen - look at him in the eye. I will find you, and count the sum of distance between each two different ‘‘cff" as substrings of the message.

 

Input

An integer T (1≤T≤100), indicating there are T test cases.
Following T lines, each line contain an integer n (3≤n≤201314), as the identifier of message.

 

Output

The output contains exactly T lines.
Each line contains an integer equaling to:

∑i<j:sn[i..i+2]=sn[j..j+2]=‘‘cff"(j−i) mod 530600414,

where sn as a string corresponding to the n-th message.

 

Sample Input

9
5
6
7
8
113
1205
199312
199401
201314

 

Sample Output

Case #1: 5
Case #2: 16
Case #3: 88
Case #4: 352
Case #5: 318505405
Case #6: 391786781
Case #7: 133875314
Case #8: 83347132
Case #9: 16520782

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  6343 6342 6341 6340 6339 

 

 

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define N 201314
 #define mod 530600414
 #define gep(i,a,b) for(int i=a;i<=b;i++)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define ll  long long
 int t,id;
 ll f[N+],c[N+],s[N+],n[N+];
 /*
 f ； 任意两个c的坐标差之和
 c :  字符串里c的个数
 s :  字符串里所有的c的坐标和
 n ； 字符串的长度
 例如 ：
 cffffcff      ffcffcffffcff
 ((8-6)+(8-1))*3
 3*2+6*2+11*2
 上面两个式子的和==f[7]
 (1+6)+(3+6+11)+8*3==s[7]
 */

 void init()
 {
     c[]=,s[]=,n[]=,f[]=;
     c[]=,s[]=,n[]=,f[]=;
     gep(i,,N){
         f[i]=(    (f[i-]+f[i-])%mod    +  (((c[i-]*n[i-]-s[i-])%mod+mod)%mod)*c[i-]%mod   +(c[i-]*s[i-]%mod)   )%mod;
         c[i]=(c[i-]+c[i-])%mod;
         n[i]=(n[i-]+n[i-])%mod;
         s[i]=((s[i-]+s[i-])%mod+(c[i-]*n[i-])%mod)%mod;
         //if(i<=12)printf("%lld %lld %lld %lld\n",f[i],c[i],n[i],s[i]);
     }
 }
 int main()
 {
     init();
     scanf("%d",&t);
     gep(i,,t){
         scanf("%d",&id);
         printf("Case #%d: %lld\n",i,f[id]);
     }
     return ;
 }